Quadratic equation using quadratic formula

Examples

Here's an example of solving a quadratic equation using the quadratic formula. Solve the equation: $2x^2 + 3x - 5 = 0$

Explanation:

Step 1: Write the equation in standard form: $2x^2 + 3x - 5 = 0$

Step 2: Identify the values of $a, b,$ and $c$:

$a = 2, b = 3,$ and $c = -5$

Step 3: Substitute the values of $a, b,$ and $c$ in the quadratic formula:

$x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x = \frac{-3 \pm\sqrt{(3)^2 - 4(2)(-5)}} { 2(2)}$

Step 4: Simplify the expression inside the square root: $(3)^2 - 4(2)(-5) = 9 + 40 = 49$

Step 5: Since the expression inside the square root is positive, the quadratic equation has real solutions.

$x = \frac{-3 \pm\sqrt{49} }{ 2(2)}$

Step 6: Simplify the expression inside the square root by finding its square root: $\sqrt{49} = 7$

$x = \frac{-3 \pm7 }{ 4}$

Step 7: Solving for $x$ to get the two possible solutions for the quadratic equation:

$x = \frac{(-3 + 7)} 4$ or $x = \frac{(-3 - 7)} 4$

Simplifying the solutions:

$x = \frac{4} 4$ or $x = \frac{-10} 4$

$x=1$ or $x = \frac{-5} 2$

Step 8: The solutions to the quadratic equation are $x = 1$ and $x = \frac{-5} 2$

Solution:

$2x^2 + 3x - 5 = 0$

Here, $a = 2, b = 3,$ and $c = -5$ Quadratic Formula: $x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute the values of $a, b,$ and $c$ in the quadratic formula:

$x = \frac{-3 \pm\sqrt{(3)^2 - 4(2)(-5)}} { 2(2)}$ $x = \frac{-3 \pm\sqrt{9 + 40}} { 2(2)}$

$x = \frac{-3 \pm\sqrt{49} }{ 2(2)}$ $x = \frac{-3 \pm7 }{ 4}$

$x = \frac{(-3 + 7)} 4$ or $x = \frac{(-3 - 7)} 4$ $x = \frac{4} 4$ or $x = \frac{-10} 4$

$x=1$ or $x = \frac{-5} 2$

The solution set of the quadratic equation is $\{1, -\frac52\}$